Question

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Answer 1

Step-by-step explanation:

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Answer 2

प्रश्न:-

1) यदि $\sf{a+\dfrac{1}{a} = \dfrac{8}{3}}$ तो $\sf{a^2+\dfrac{1}{a^2} = ?}$

2) यदि $\sf{x+\dfrac{1}{x} = \dfrac{4}{5}}$ तो $\sf{x^2+\dfrac{1}{x^2} = ?}$

3) यदि $\sf{a+\dfrac{1}{a} = \dfrac{7}{3}}$ तो $\sf{a^2 + \dfrac{1}{x^2} = ?}$

4) यदि $\sf{x+\dfrac{1}{x} = 20}$ तो $\sf{x^2 + \dfrac{1}{x^2} = ?}$

5) यदि $\sf{y + \dfrac{1}{y} = \dfrac{8}{5}}$ तो $\sf{y^2 + \dfrac{1}{y^2} = ?}$

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हल:-

1) प्रश्न में दिया हुआ:-

$\sf{a + \dfrac{1}{a} = \dfrac{8}{3}}$

आवश्यकता:-

$\sf{a^2+\dfrac{1}{a^2}}$ का मूल।

हल:-

$\sf{a+\dfrac{1}{a} = \dfrac{8}{3}}$

बराबर के दोनों बगल वर्ग करते हुए,

$\sf{\bigg(a +\dfrac{1}{a}\bigg)^2 = \bigg(\dfrac{8}{3}\bigg)^2}$

= $\sf{a^2 + 2\times a\times \dfrac{1}{a} + \dfrac{1}{a^2} = \dfrac{64}{9}}$

= $\sf{a^2 + 2 + \dfrac{1}{a^2} = \dfrac{64}{9} }$

= $\sf{a^2 + \dfrac{1}{a^2} = \dfrac{64}{9} - 2}$

= $\sf{a^2 + \dfrac{1}{a^2} = \dfrac{64-18}{9}}$

= $\sf{a^2 + \dfrac{1}{a^2} = \dfrac{46}{9}}$

$\sf{\therefore a^2 +\dfrac{1}{a^2}}$ का मूल है $\sf{\dfrac{46}{9}}$

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2) प्रश्न में दिया हुआ:-

$\sf{x+\dfrac{1}{x} = \dfrac{4}{5}}$

आवश्यकता:-

$\sf{x^2 + \dfrac{1}{x^2}}$ का मूल।

हल:-

$\sf{x+\dfrac{1}{x} = \dfrac{4}{5}}$

बराबर के दोनों बगल वर्ग करते हुए,

$\sf{\bigg(x+\dfrac{1}{x}\bigg)^2 = \bigg(\dfrac{4}{5}\bigg)^2}$

= $\sf{x^2 + 2\times x \times \dfrac{1}{x} + \dfrac{1}{x^2} = \dfrac{16}{25}}$

= $\sf{x^2 + 2 + \dfrac{1}{x^2} = \dfrac{16}{25}}$

= $\sf{x^2 + \dfrac{1}{x^2} = \dfrac{16}{25} - 2}$

= $\sf{x^2 + \dfrac{1}{x^2} = \dfrac{16-50}{25}}$

= $\sf{x^2 + \dfrac{1}{x^2} = \dfrac{-34}{25}}$

$\sf{\therefore x^2+\dfrac{1}{x^2}}$ का मूल है $\sf{\dfrac{-34}{25}}$

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3) प्रश्न में दिया हुआ:-

$\sf{a+\dfrac{1}{a} = \dfrac{7}{3}}$

आवश्यकता:-

$\sf{a^2+\dfrac{1}{a^2}}$ का मूल।

हल:-

$\sf{a+\dfrac{1}{a} = \dfrac{7}{3}}$

बराबर के दोनों बगल वर्ग करते हुए,

$\sf{\bigg(a+\dfrac{1}{a}\bigg)^2 = \bigg(\dfrac{7}{3}\bigg)^2}$

= $\sf{a^2 + 2\times a\times \dfrac{1}{a} + \dfrac{1}{a^2} = \dfrac{49}{3}}$

= $\sf{a^2 + 2+ \dfrac{1}{a^2} = \dfrac{49}{9}}$

= $\sf{a^2 + \dfrac{1}{a^2} = \dfrac{49}{9} - 2}$

= $\sf{a^2 + \dfrac{1}{a^2} = \dfrac{49-18}{9}}$

= $\sf{a^2 + \dfrac{1}{a^2} = \dfrac{31}{9}}$

$\sf{\therefore a^2 + \dfrac{1}{a^2}}$ का मूल है $\sf{\dfrac{31}{9}}$

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4) प्रश्न में दिया हुआ:-

$\sf{x+\dfrac{1}{x} = 20}$

आवश्यकता:-

$\sf{x^2 + \dfrac{1}{x^2}}$ का मूल।

हल:-

$\sf{x+\dfrac{1}{x} = 20}$

बराबर के दोनों बगल वर्ग करते हुए,

$\sf{\bigg(x+\dfrac{1}{x}\bigg)^2 = (20)^2}$

= $\sf{x^2 + 2\times x \times \dfrac{1}{x} + \dfrac{1}{x^2} = 400}$

= $\sf{x^2 + 2 + \dfrac{1}{x^2} = 400}$

= $\sf{x^2 + \dfrac{1}{x^2} = 400-2}$

= $\sf{x^2 + \dfrac{1}{x^2} = 398}$

$\sf{\therefore x^2+\dfrac{1}{x^2}}$ का मूल है 398.

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5) प्रश्न में दिया हुआ:-

$\sf{y + \dfrac{1}{y} = \dfrac{8}{5}}$

आवश्यकता:-

$\sf{y^2 + \dfrac{1}{y^2}}$ का मूल।

हल:-

$\sf{y + \dfrac{1}{y} = \dfrac{8}{5}}$

बराबर के दोनों बगल वर्ग करते हुए,

$\sf{\bigg(y+\dfrac{1}{y}\bigg)^2 = \bigg(\dfrac{8}{5}\bigg)^2}$

= $\sf{y^2 + 2\times y \times \dfrac{1}{y} + \dfrac{1}{y^2} = \dfrac{64}{25}}$

= $\sf{y^2 + 2 + \dfrac{1}{y^2} = \dfrac{64}{25}}$

= $\sf{y^2 + \dfrac{1}{y^2} = \dfrac{64}{25} - 2}$

= $\sf{y^2 + \dfrac{1}{y^2} = \dfrac{64-50}{25}}$

= $\sf{y^2 + \dfrac{1}{y^2} = \dfrac{14}{25}}$

$\sf{\therefore y^2 + \dfrac{1}{y^2}}$ का मूल है $\sf{\dfrac{14}{25}}$

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